118 ANALYTICAL MECHANICS the frictional component of R by F and the normal component by N, and equating the components of the kinetic reaction along and at right angles to the plane to the sums of the corresponding components of the forces we obtain dv - n ^\ m— = <wg sm a — F, (a; at 0 = N — mg cos a. (b) But if fjL is the coefficient of friction then F=vN [p. 22] = fimgcosa [by (b)]. Substituting this value of F in equation (a) we obtain dv m— = mg sin a — /* mg cos a, at or -77 = g (sin a - p cos a). at Thus the acceleration is constant. Therefore the equations of motion are obtained by replacing fbyg (sin a — /* cos a) in equations (1) to (3) of page 113: v = v0 + gt (sin a — /* cos a), s — vQt +1 gtf2 (sin a — /z cos a), fl2 = 2;02+ 2#$ (sin a — /* cos a). PROBLEMS. 1. A car weighing 10 tons becomes uncoupled from a train which is moving down a grade of 1 in 200 at the rate of 50 miles per hour. If the frictional resistance is 15 pounds per ton, find the distance the car will travel before coming to rest. 2. The pull of a locomotive is 2500 pounds. Find the velocity it can give in 5 minutes to a train which weighs 75 tons. Take 10 pounds per ton for the resistance and consider the tracks to be horizontal. 3. In the preceding problem suppose the tracks to have a grade of 1 in 200 and find the velocity (a) going down grade and (b) going up grade.